Question: Subtract the following rational expressions. $\dfrac{6z}{6z+1}-\dfrac{2z^2}{3z+4}=$
Explanation: We can subtract two rational expressions whose denominators are equal by subtracting the numerators and keeping the denominator the same. [Does this fit with how we subtract rational numbers?] When the denominators are not the same, we must manipulate them so that they become the same. In other words, we must find a common denominator. Since the two denominators do not share any common factors, the common denominator is simply the product of these two denominators: $({6z+1})\cdot({3z+4})$. Let's manipulate the expressions to have that denominator: $\begin{aligned} &\phantom{=}\dfrac{6z}{{6z+1}}-\dfrac{2z^2}{{3z+4}} \\\\ &=\dfrac{6z\cdot({3z+4})}{({6z+1})\cdot({3z+4})}-\dfrac{2z^2\cdot({6z+1})}{({3z+4})\cdot({6z+1})} \end{aligned}$ [Why did we do that?] Now that both denominators are the same, let's subtract! $\begin{aligned} &\phantom{=}\dfrac{6z\cdot(3z+4)}{(6z+1)\cdot(3z+4)}-\dfrac{2z^2\cdot(6z+1)}{(3z+4)\cdot(6z+1)} \\\\ &=\dfrac{6z\cdot(3z+4)-2z^2\cdot(6z+1)}{(6z+1)(3z+4)} \\\\ &=\dfrac{18z^2+24z-12z^3-2z^2}{(6z+1)(3z+4)} \\\\ &=\dfrac{-12z^3+16z^2+24z}{(6z+1)(3z+4)} \end{aligned}$ In conclusion, $\dfrac{6z}{6z+1}-\dfrac{2z^2}{3z+4}=\dfrac{-12z^3+16z^2+24z}{(6z+1)(3z+4)}$